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Contribution of the auto dissociation of water

  1. May 10, 2015 #1
    1. The problem statement, all variables and given/known data
    How much does the auto - dissociation of water contribute to the pH a 10e+3M aqueous solution of sulphuric acid?

    2. Relevant equations
    Ka = [H30+][SO42-}/[HSO4-] = 1.1e-2
    Kw = [OH-][H30+]

    3. The attempt at a solution
    Sulphuric acid is a strong acid in the first hydrolysis step:
    H2SO4 + H20 -> H30+ + HSO4-
    Sulphuric acid will dissociate completely to form 10^3M of both H30+ and HSO4-
    I also neglected the auto dissociation of water, which i will come back to later.

    Leaving the first hydrolysis step aside, looking at the second hydrolysis step:
    HSO4- + H20 <=> SO42- + H30+
    We know that the initial concentration of HSO4- is 10^3M from the first hydrolysis step.
    400*Ka = 4.4 >> 10^3, so the equilibrium concentration of HSO4- is not equal to the initial concentration.
    so [HSO4-] = initial [HSO4-] - [SO42-]formed
    the concentration of h30+ formed from the second step is equal to that of so42-: [H3O+] = [S042-]

    combining the above facts, i got a quadratic equation linking Ka and [H30+]
    let x = [H30+]
    x^2 + xKa - Ka[HSO4-] = 0
    Solving this gives x = 9.226e-4, so [H30+] formed from second hydrolysis step is 9.226e-4M (the other root is negative and is ignored)

    As the [H30+] formed from the second hydrolysis step (9.226e-4 M) is quite close to the [H30+] formed from the first step (1.0 e-3M), we cannot assume that the total [H30+] is governed by the first dissociation step only. The total [H30+] is the sum of the [H30+] from both the 1st step and the 2nd step, giving [H30+] = 1.923e-3M

    To find the effect of the auto dissociation of water, I used the Kw relation.
    Kw = [OH-][H30+] , where [OH-] and [H30+] refer to the total concentrations present in the solution.
    Water is the only species that will dissociate and form OH- ions, so the total [OH-] in the solution is equal to the [OH-] from the dissociation of water, which is also equal to the [h30+] formed from the dissociation of water.

    Knowing this, we find [h30+] (from water) = [OH-] = Kw/[H30+],
    giving the h30+ contribution of the auto dissociation of water to be 5.2e-12M

    however the answer is 5.43e-12M.

    What have I done wrong?
    Thank you very much in advance.
     
  2. jcsd
  3. May 10, 2015 #2

    Borek

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    Pure sulfuric acid is 18.7 M, 1000 M solution is completely unreasonable.

    No hydrolysis here, only dissociation.

    Please elaborate on this step.
     
  4. May 10, 2015 #3
    Yes typo, it should be 10e-3.

    Ka = [H30+]^2/([HSO4-] - [H30+])
    rearranging and letting x be [H30+] gives the quadratic equation x^2 + xKa - Ka[HSO4-] = 0.
    Well, you solve the quadratic by plugging into a calculator?
    x^2 + 0.011x - 1.e-5 = 0 giving x = 9.226e-4
    My calculator has a function that solves quadratic equations so......
     
  5. May 10, 2015 #4

    mfb

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  6. May 10, 2015 #5
    Which part are you referring to?

    I just used matlab to solve the quadratic and still got the same answer for [H3O+].
     
  7. May 10, 2015 #6

    mfb

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    The part where x=8.44*10-4 is given as solution, which does not agree with your result.
     
  8. May 10, 2015 #7
    Sorry, there was another typo
    the quadratic should be x^2 + 0.011*x - 0.000011.
    which still gives x = 0.0009226.
    Thank you very much.
     
  9. May 10, 2015 #8

    Borek

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    Doesn't look OK to me. Try to construct the ICE table.
     
  10. May 10, 2015 #9
    Can you just tell me what's wrong with it?

    initial: [HSO4-] = 0.001M, [H30+] (ignoring what was formed from the first dissociation) = 0M, [SO42-] = 0M
    change: [HSO4-] = -x, [H30+] = +x, [SO42-] = +x
    equilibrium: [HSO4-] = 0.001-x, [H30+] = x, [SO42-] = x
     
  11. May 10, 2015 #10

    Borek

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    And how does the dissociation know part of the identical ions present in the solution have to be ignored?
     
  12. May 11, 2015 #11
    oh....
    so the [H30+] should be the total concentration in the solution,
    initial: [HSO4-] = 0.001M, [H30+] = 0.001M, [SO42-] = 0M
    change: [HSO4-] = -x, [H30+] = +x, [SO42-] = +x
    equilibrium: [HSO4-] = 0.001-x, [H30+] = 0.001+x, [SO42-] = x

    giving the quadratic x^2 + 0.012x -0.000011=0, so x = 8.56e-4
    so the total [H30+] is 0.001856M, and following the steps from before i get 5.39e-12M [h30+] from the dissociation of water?
     
  13. May 11, 2015 #12

    Borek

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    These numbers look much better.
     
  14. May 12, 2015 #13

    James Pelezo

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    Isn't the 5.39 x 10-12 the hydroxide ion concentration since [H3O+] = 0.001856M? From Kw = [H3O+][OH-] => 1 x 10-14 = (0.001856)[OH-] => [OH-] = (1x10-14/0.001856)M = 5.39 x 10-12M. You have [H3O+] after the 5.39 x 10-12M.
     
  15. May 13, 2015 #14

    Borek

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    Concentration of OH- from water autodissociation equals concentration of H+ from water autodissociation, so yes, 5.39×10-12 M is both.

    And that's an interesting observation. In this case - as the contribution of H+ from water autodissociation is many orders of magnitude lower than the total concentration of H+ - this approach (calculating OH- and ignoring changes in H+) is much faster.
     
  16. May 13, 2015 #15

    James Pelezo

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    Interesting problem when using HSO4-. One has to use the quadratic b/c of the high ionization of the second ionization step. Most problems are weak monoprotic that ionize very little. It is a refreshing point of issue... Thanks
     
  17. May 14, 2015 #16

    epenguin

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    Well. o_O Well. Hmphhh.

    I'd rather say you have to use a quadratic because the concentration of the acid was within about one order of magnitude of the Ka. Any farther than that either way and you wouldn't need to solve quadratics. And even in this case note that the difference from the straight off approximation of [H+] = 2×10-3 is not very great.

    As long as you realise this more important approximation, to be able to handle a problem like this is useful; the approach appears to be correct (I have not checked or calculated myself the result - as the OP has an unconcluded thread open I had no evidence that such investment of my time is useful, see my sig.).

    Although that part of the problem is useful excercise I hardly think the final object of the excercise, namely to determine the minuscule number of water moles contributing as asked let alone the even more minuscule difference between an exact or approximate calculation of them, is useful. It is one of several rather strange perplexing questions that barely make sense we have had recently. I think it would be good to know who asks them. (Though I have a general idea).
     
    Last edited: May 14, 2015
  18. May 14, 2015 #17

    James Pelezo

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    Just a little FYI if interested... a simplifying assumption for when to use the quadratic or not use the quadratic for solving weak acid equilibria problems is to check ... if [Molar Concentration of Acid/Ka-value] > 100 then the quadratic is not used. If [C]/Ka < 100, ionization is high enough to affect equilibrium concentrations and should be used to determine equilibrium concentrations of the acid of interest. Have a good day.:wink:
     
  19. May 15, 2015 #18

    Borek

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